\documentclass[UTF8]{article}
\usepackage{amsmath}
 \usepackage{CJK}
\author{lijiguo {   }{  } 201618013229046}
\title{assignment for stachostics}
\begin{document}
  \maketitle
  测试中文
  \section{learn to use Tex}
  There is no doubt that Tex is very important for us students to write a paper, but unfortunatelly, I am new to this powerful tool.
  So, the first task for me is learn to use it. \newline
  Up to now, I can not still to input a Chinese, but I am improving.

  \section{a question about random event}
  \emph{Q:} if $P(A)>0$, and $P(B|A)=P(B|\bar{A})$
  \newline try to demonstrate that: $\boldsymbol{A} \boldsymbol{is} \boldsymbol{independent} \boldsymbol{with} \boldsymbol{B}$.
  \newline
  \newline \emph{S:} according to the bayes theorem, we know that
  \newline $P(B|A)=\frac{P(AB)}{P(A)}$
  \newline $P(B|\bar{A})=\frac{P(\bar{A} B)}{P(B)}$
  \newline So we get that:
  \newline $P(\bar{A})P(AB)=P(A)P(\bar{A}B)$
  \newline As we know:
  \newline $P(\bar{A})=1-P(A)$
  \newline $P(\bar{A}B)=P(B)-P(AB)$
  \newline We can get that:
  \newline $[1-P(A)]P(AB)=P(A)[P(B)-P(AB)]$
  \newline $P(AB)=P(A)P(B)$
  \newline We can get that $\boldsymbol{A} \boldsymbol{is} \boldsymbol{independent} \boldsymbol{with} \boldsymbol{B} from this equation$.
  \section{a question for continuous random variable}
  \emph{Q:} if the probability density function (pdf) of $X$ is $f_X(x)$
  \newline give the pdf of $Y=X^2$
  \newline
  \newline \emph{S:} we can easily get that it is impossible for $Y$ to be a negative, so
  \newline $F_Y(y)=0 \qquad if\quad y\le 0$
  \newline when $y \ge 0$
  \newline $F_Y(y)=P(Y\le y)=P(X^2 \le y)=P(-\sqrt{y}\le X \le \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})$
  \newline so we can get the probability density function of Y is:
  \newline
  $ f_Y(y) = F'_Y(y)=
  \begin{cases}
    \frac{1}{2\sqrt{y}}[f_X(\sqrt{y})+f_X(-\sqrt{y})],\\
    1,
  \end{cases}$

\end{document}
